i tried to do like dis plz chk is it rite or wrong.
[2(4(327) + 495 ]- [ 2(4(327) + 523 ]
= -28
Bond enthalpy is the energy required to break a mole of a certain type of bond.
O=O = 495 kj/mol
S-F = 327 kj/mol
S=O = 523 kj/mol
Use average bond enthalpies to estimate the enthalpy delta H (rxn) of the following reaction:
2SF4 + O2 ---- 2OSF4
Express your answer numerically in kilojoules.
3 answers
i got it .. no worries .. answer was -551 thnks anywyz .. DR BOB FOR this question .. i asked so many times.. so really SORRRY DR BOB >. to bother u lot..
-551 kJ is what I obtained also.
S-F = 327*4 = 1308
2mols of that so 2*1308 = 2616
+ O=O = 495.
2616+495 = 3111 for reactants.
Products:
4 S-F bonds @ 327 each = 1308
+ 1 S=O @ 523 = 1831 for each molecule.
There are two of them so 2*1831= 3662.
Then reactants - products = 3111-3662 = -551 kJ.
S-F = 327*4 = 1308
2mols of that so 2*1308 = 2616
+ O=O = 495.
2616+495 = 3111 for reactants.
Products:
4 S-F bonds @ 327 each = 1308
+ 1 S=O @ 523 = 1831 for each molecule.
There are two of them so 2*1831= 3662.
Then reactants - products = 3111-3662 = -551 kJ.
Maybe chemistry is your profession cuz dude, you can't spell worth s***
0 answers